What is the extraneous solution to these equations? $\dfrac{x^2 + 6x}{x + 5} = \dfrac{-3x - 20}{x + 5}$
Multiply both sides by $x + 5$ $ \dfrac{x^2 + 6x}{x + 5} (x + 5) = \dfrac{-3x - 20}{x + 5} (x + 5)$ $ x^2 + 6x = -3x - 20$ Subtract $-3x - 20$ from both sides: $ x^2 + 6x - (-3x - 20) = -3x - 20 - (-3x - 20)$ $ x^2 + 6x + 3x + 20 = 0$ $ x^2 + 9x + 20 = 0$ Factor the expression: $ (x + 4)(x + 5) = 0$ Therefore $x = -4$ or $x = -5$ At $x = -5$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -5$, it is an extraneous solution.